## Devlin's Angle |

For instance, since the publication of my book *The Math Gene* in 2000, I've received a steady stream of messages from readers who believe there is an error in the proof I give there that there are infinitely many primes. There isn't. The confusion that arises in some readers'
minds comes from my attempt to lay out the standard argument in a completely linear fashion. When I wrote the book, I formulated the proof the way I did because experience had shown me that an argument by cases can cause problems for some people.
Unfortunately, eliminating the cases
approach results in a different problem - rather, two different problems, as it turned out.

The argument by cases will be familiar to most readers of this column. The idea is to show that if you start to list the primes in order, P(1),P(2), P(3), etc., then the list continues for ever. To show this, you assume you have reached some stage P(N), and you prove that you can always find at least one more prime with which to continue the list. The idea - and it's a brilliant one, known to the ancient Greeks and described in Euclid's book *Elements*
- is to consider the number

Q = [P(1) x P(2) x ... x P(N)] + 1obtained by multiplying together all the primes listed so far and then adding 1 to the product. Now the argument splits into two cases. If Q is prime, then since Q is bigger than P(N), it is a prime not already listed. If Q is not prime, then there must be some prime P that divides into Q. But P cannot be any one of P(1), ..., P(N), since dividing any of those into Q leaves a remainder of 1. So P is a prime not already listed. Either way, there is a prime not in the list, so the list can be continued.

Now, if you are one of the many people who finds the above argument unproblematic, you may have trouble understanding why anyone would object to it. But believe me, many people do. (One problem that some people have - a problem I hoped to avoid by organizing the argument differently - is to assume that we have in some way identified the next prime to be added to the list. We have not; we've just showed that there is at least one more prime out there, and hence the list may be continued by picking the next one, *whatever it may be.*)

Knowing that the above argument causes some people difficulties, when I wrote *The Math Gene,* I formulated the proof in a slightly different way, as a proof by contradiction. To whit:

Suppose that there are only finitely many primes. List them all as P(1), P(2), ..., P(N). (Thus P(N) is the largest prime.) Now look at the number

Q = [P(1) x P(2) x ... x P(N)] + 1Next we show that Q is prime. To do this, we show that it has no prime divisors. Well, all primes are in the list P(1), ..., P(N), and each of those numbers leaves a remainder of 1 when you divide it into Q. Hence Q can have no prime divisors. Hence Q must itself be prime.

But Q is bigger than P(N), and P(N) is supposedly the largest prime, so we have a contradiction. Hence our initial assumption is false, which is to say, there must be infinitely many primes.

Now you might suspect that the reason some people have problems with this argument is that it uses the method of contradiction, and for some of my correspondents that seems to have been the case. But I also get letters from readers who clearly understand how proof by contradiction works, but who nevertheless feel the proof is wrong. Their misconception almost always comes from believing the argument claims that the number

Q = [P(1) x P(2) x ... x P(N)] + 1is prime

[2 x 3 x 5 x 7 x 11 x 13] + 1 = 30031 = 59 x 509But the argument I give makes no such claim. The claim that "Q is prime" is made for a specific (and decidedly hypothetical) value of N, namely the N for which P(N) is the supposed largest prime number, and the claim depends on that assumed property of P(N).

Now, I might have avoided problems if I had phrased the last part of my proof to avoid using the phrase "Q is prime". This can easily be done. I didn't because I wanted to split the entire proof into small, easily digested pieces, but maybe that was a poor choice. My choice did, however, highlight a problem that is far more fundamental. After engaging in several exchanges with readers who had trouble with my proof, I eventually realized that the problem was, at heart, confusion between variables and fixed, but unknown, constants. (Actually, in the case of my primes proof, a constant that does not exist at all outside the context of the proof!) And here is the lesson to be learned, I think. We mathematicians happily use letters to denote both variables and constants. But many people don't see the distinction. They see a formula like

Q = [P(1) x P(2) x ... x P(N)] + 1and because N is a letter and not a specific number, such as 3, 6, or 29, they assume it denotes ANY number.

I suspect, but do not know for certain, that computer programmers would not make such an error. They, after all, have to know that once you set a variable X equal to a specific numerical value in a program, then it denotes that specific value, and nothing else, until reset to another specific value. They understand too the difference between a free variable and a variable that is bound by a context, such as being introduced in a subroutine. But I suspect that many students pass through one or more college level mathematics courses without ever coming to grips with the way mathematicians use letters to denote any one of a variable, a fixed but unknown number, or a known specific number.

The second matter that regularly lands upon my desk - or in my email inbox - is occasioned by attempts (in my own writing and that by others) to try to explain the basic ideas of the topological study of surfaces (specifically, orientability) by distinguishing between a wedding band (or
cylinder) and a Moebius strip, in terms of the former "having two sides" and the latter "one side". In my book
*Mathematics: The New Golden Age,* for example, I try to explain orientability by considering what happens when you take a closed loop with an arrowhead in the surface and sliding it around, whereupon, in the case of a Moebius band, the arrow changes from clockwise to counterclockwise when you return to the starting point - something that does not happen in the case of a cylindrical (wedding) band. I illustrate the point with the following diagram. (The loop starts just to the left of the line and moves around the band in a clockwise direction until it arrives back at the starting line.)

The third confusion that I get asked to clear up (or accused - often with great vehemence - of getting wrong in my writing) concerns probability. But I've already used up my monthly allotment of space, so I'll turn to that in a later column. Stay tuned.

Devlin's Angle is updated at the beginning of each month.